Question: The lifespans of turtles in a particular zoo are normally distributed. The average turtle lives $105$ years; the standard deviation is $11.8$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a turtle living less than $140.4$ years.
Solution: $105$ $93.2$ $116.8$ $81.4$ $128.6$ $69.6$ $140.4$ $99.7\%$ $0.15\%$ $0.15\%$ We know the lifespans are normally distributed with an average lifespan of $105$ years. We know the standard deviation is $11.8$ years, so one standard deviation below the mean is $93.2$ years and one standard deviation above the mean is $116.8$ years. Two standard deviations below the mean is $81.4$ years and two standard deviations above the mean is $128.6$ years. Three standard deviations below the mean is $69.6$ years and three standard deviations above the mean is $140.4$ years. We are interested in the probability of a turtle living less than $140.4$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $99.7\%$ of the turtles will have lifespans within 3 standard deviations of the average lifespan. The remaining $0.3\%$ of the turtles will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({0.15\%})$ will live less than $69.6$ years and the other half $({0.15\%})$ will live longer than $140.4$ years. The probability of a particular turtle living less than $140.4$ years is ${99.7\%} + {0.15\%}$, or $99.85\%$.